Electron Speed?

Question

An electron starts from rest 74.5 cm from a fixed point charge with Q = -0.175 µC. How fast will the electron be moving when it is very far away?

Answer 1

The field potential from a point charge is

V(r)= q1/(4π*e0*r)

The potential energy of charge q2 a distance d from the point charge is

PE = V(d)*q2 = q1*q2/(4π*e0*d)

very far away, PE goes to zero and is converted to kinetic energy of the electron which would be

PE(d)= .5*me*ve^2, where me is the mass of the electron.

Then ve = √[2*PE(d)/me] = √[2*q1*qe/(4π*e0*d)*me]

me = 9.1*10^-31 kg
qe = -1.6*10^-19 coul
e0 = 8.9*10^-12 Fd/m

Answer 2

F=Qq/(4pi epsilon zero r squared) so a=Qq/(4pi epsilon zero r squared m). So to get the velocity at infinity you integrate this from 0.745 to infinity. This gives you Qq/ (4pi epsilon zero R m). Q is 0.175 microcoulombs, q is the charge on an electron, pi is 3.14159, epsilon zero is the electric permittivity of a vacuum, r is the distance of the electron from the fixed point charge, which varies, R is 0.745 metres, m is the mass of the electron. Substitute these values into Qq/(4pi epsilon zero Rm) and you'll get the speed of the electron in metres/sec when it is far away. It has a finite value less than the speed of light.

Answer 3

For a point charge, electric potential nearby = V = kQ/r

qVf-qVo=KEf-KEo
-1.6x10^-19(kQ/HUGE - kQ/.745) = 1/2(9.11x10^-31)vf^2 - 0

kQ/Huge ~0.
K=9x10^9

And solve for vf, final velocity.

Answer 4

As far as I know it will be traveling at the speed of light.